Solving Systems Consisting of Linear and Quadratic Equations SAT Notes

A quadratic system contains a quadratic equation with another quadratic or linear equation in the same variables. A solution is an ordered pair that satisfies both equations, so graphically it is an intersection point.

Zero, one, or two line-parabola intersections

How a line can meet a parabola

The three panels use exact equations to show a line missing, touching, or crossing the same parabola.

No real intersection

One tangent intersection

Two real intersections

Algebra and graph agreement after substitution
Reduced discriminant	Real ordered pairs	Graph
$$D<0$$	Zero	The line misses the parabola.
$$D=0$$	One	The line is tangent to the parabola.
$$D>0$$	Two	The line crosses the parabola twice.

Solve a line-parabola system by substitution

Isolate
Write one equation as $y=\text{expression}$ when necessary.
Substitute
Replace $$y$$ in the other equation with that expression.
Solve
Solve the resulting one-variable quadratic and keep every valid root.
Recover coordinates
Substitute each input back to find its corresponding output.
Verify
Check each ordered pair in both original equations.

Worked example

Find two exact intersections

Solve $$y=x^2-1$$ and $$y=x+1$$ .

Set expressions equal
$$x^2-1=x+1$$ , so $$x^2-x-2=0$$ .
Factor
$$(x-2)(x+1)=0$$ , giving $$x=2$$ or $$x=-1$$ .
Find outputs
From $$y=x+1$$ , the outputs are $$3$$ and $$0$$ .
Verify
Both $$(2,3)$$ and $$(-1,0)$$ satisfy both original equations.

The system solutions are

\((-1,0)\)

and

\((2,3)\)

A line and parabola with two exact solutions

Quadratic plus quadratic

Set the two function expressions equal, simplify, solve the resulting equation, then recover the output coordinate. At this chapter's level the reduced equation remains manageable; no quartic technique is required.

Worked example

Intersect two parabolas

Solve $$y=x^2-1$$ and $$y=-x^2+7$$ .

Set equal
$$x^2-1=-x^2+7$$ .
Solve
$$2x^2=8$$ , so $x=\pm2$ .
Recover output
For either input, $$y=x^2-1=3$$ .

The intersections are

\((-2,3)\)

and

\((2,3)\)

Two quadratic graphs with exact intersections

Parameters and tangency

Worked example

Choose a tangent horizontal line

For what value of $$k$$ is $$y=k$$ tangent to $$y=x^2-6x+13$$ ?

Substitute
$$x^2-6x+13=k$$ , or $$x^2-6x+(13-k)=0$$ .
Set the discriminant
$$D=(-6)^2-4(1)(13-k)=4k-16$$ .
Tangency
$$D=0$$ gives $$4k-16=0$$ .

\(k=4\)

, matching the parabola's minimum output.

Common mistakes

Do not report input values alone when ordered pairs are requested.
Substitute every quadratic root back; different inputs can produce different outputs.
Use the discriminant only after the system has been reduced to one quadratic in one variable.
Do not assume two graphs must intersect.
For a contextual system, test any domain restrictions after algebraic verification.

Mini check

Interpret a tangent

If substitution produces $$x^2-8x+16=0$$ , how many real intersection points does the original line-parabola system have?

Zero
One
Two
Four

Show answer and explanation

Answer: One

The quadratic is $$(x-4)^2=0$$ , so it has one repeated root and the graphs are tangent.

Key takeaways

What to remember

System solutions are intersection points and must satisfy both equations.
Substitution converts a line-parabola system into a quadratic equation.
The reduced discriminant predicts zero, one, or two real line-parabola intersections.
Always recover and verify the output coordinate for every input solution.

Continue learning

Put these notes into practice

Apply the ideas with SAT-style questions, then reinforce key details with flashcards.

Go to Practice Review Flashcards

Reduced discriminant	Real ordered pairs	Graph
$\(D<0\)$	Zero	The line misses the parabola.
$\(D=0\)$	One	The line is tangent to the parabola.
$\(D>0\)$	Two	The line crosses the parabola twice.